For what value(s) of k is the line x+2y+k=0 tangent to the circle x^2+y^2-2x+4y+1=0

It's an analytic geometry question. Particularly about circles.
Also, this is another question:

Find an equation of the circle through points (1,5), (-2,3), and (2,-1).
The desired equation is:
x^2 + y^2 + D'x + E'y + F' = 0

Here is my solution:
1 + 25 + D' + 5E' + F' = 0
4 + 9 - 2D' + 3E' + F' = 0
4 + 1 + 2D' - E' + F' = 0

then, simplifying it:
D' + 5E' + F' = -26
-2D' + 3E' + F' = -13
2D' - E' + F' = -5

NOW, HERE'S THE TRICKY PART:
My professor then said to solve it simultaneously.
Now, I don't know how to do that. She didn't even taught us how to do it.
She just wrote on the whiteboard:

D' = -9/5
E' = -19/5
F' - -26/5

Anyone know how she came up with those answers for D', E', and F'?

I have a quiz on thursday, and I don't know how to solve simultaneously.
Help would be highly appreciated.

Cheers!

x^2 + y^2 - 2x + 4y + 1 = 0,
=> (x - 1)^2 + (y + 2)^2 = 2^2
=> center of the circle is C (1, - 2) and radius r = 2

For the line, x + 2y + k = 0, to be a tangent to the circle,
perpendicular distance from the center of the circle = radius

=> l 1 - 4 + k l / √5 = 2
=> (k - 3)^2 = 20
=> k - 3 = ± 2√5
=> k = 3 ± 2√5.

For figure, refer to the following Wolfram Alpha link:
http://www.wolframalpha.com/input/?i=x%5…

Second Problem:

D' + 5E' + F' = -26 ... ( 1 )
-2D' + 3E' + F' = -13... ( 2 )
2D' - E' + F' = -5 ... ( 3 )

(1 ) - ( 2 )
=> 3D' + 2E' = - 13 ... ( 4 )

( 3 ) - ( 1 )
=> D' - 6E' = 21 ... ( 5 )

Multiply ( 4 ) by 3 and add to ( 5 )
=> 10D' = - 18 => D' = - 9/5

Plugging ( - 9/5) in ( 4 ),